Set forms a basis, and span help

[billli]

Homework Statement

http://img16.imageshack.us/img16/6606/50381320.jpg

Homework Equations

Please see above picture

The Attempt at a Solution

I believe for question a) I just need to add up all the matrices and then row reduce to RREF, which gives me:
[1,0,0]
[0,1,0]

or Do I row reduce each matrices?

I'm really not sure how to do b), any tips would be great!

Thanks

 

[lanedance]

think about linear combinations...

in a) can every N$$\in$$ M₂₃ be written as a linear combination of the matricies given?

you could also think about what a simpler basis for M₂₃ is and how they're related

similarly in b) can the equation be written as a linear combination of the basis equations?

 

[Unco]

I only wish to elaborate on Lanedance's post with a systematic spin.

The Attempt at a Solution

I believe for question a) I just need to add up all the matrices and then row reduce to RREF, which gives me:
[1,0,0]
[0,1,0]

or Do I row reduce each matrices? Certainly not![/color]

Think of the matrices in the set S as vectors in the vector space $$M_{23}$$ (the set of all 2x3 matrices). A basis for $$M_{23}$$ is
$$\mathbf{e_1} = \begin{pmatrix}1 & 0 & 0\\ 0&0&0\end{pmatrix},$$ $$\mathbf{e_2} = \begin{pmatrix}0 & 1 & 0\\ 0&0&0\end{pmatrix}, \ldots,$$ $$\mathbf{e_6} = \begin{pmatrix}0 & 0 & 0\\ 0&0&1\end{pmatrix}.$$

The first vector in S can then be written as $$\mathbf{v_1} = [1 \; 2 \; 3 \; 4 \; 5 \; 6]^T$$ with respect to this basis, and similarly for the others, e.g., $$\mathbf{v_6} = [12 \; 10 \; 12 \; 12 \; 20 \; 18]^T$$.

So to systematically determine whether the vectors in S are linearly independent (and hence form a basis), i.e., to determine whether the equation $$k_1\mathbf{v_1} + \ldots + k_6\mathbf{v_6}=\mathbf{0}$$ has a nontrivial solution, one can row reduce the corresponding matrix $$[\mathbf{v_1} \ldots \mathbf{v_6}]$$.

I'm really not sure how to do b), any tips would be great!

It suffices to show that S spans the set of all such (i.e., of degree 4 or less) polynomials, which amounts to determining whether p1, p2, ..., p5 are linearly independent. Taking e1 = 1, e2 = x, ..., e5 = x^4, this is just like the part (a).

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Of course, in part (a) for example, if you can spot a non-trivial linear combination of the $$\mathbf{v_i}'s$$ that sums to $$\mathbf{0}$$, you are done. (I don't think it will be helpful if someone gives you one straight off the bat.)

 

[lanedance]

if you do look at the large matrix suggested by unco, its also worth considering the determinant and how it relates to linear independence and row reduction

 

[billli]

http://img239.imageshack.us/img239/6250/picture2.png
Sorry, but I do not know how to use Tex. And thanks for the tips!

From this, a) does not form the basis, and for b) p(x) belongs to the span (not too sure about b) as it has a row of zeros)

 

[Unco]

From this, a) does not form the basis,

Right,though the last column of zeros is superfluous.

and for b) p(x) belongs to the span (not too sure about b) as it has a row of zeros)

This time taking vectors in row form (which is fine), the matrix you have here (albeit again with superfluous column of zeros) is what you would set up to see if {p, p1, p2, ..., p5} is linearly independent. You have shown they are not. But this does not necessarily mean p is a linear combination of p1, ... p5. For example, consider the set {(1,0), (2,0), (0,1)}: it is linearly dependent but (0,1) is not a linear combination of (1,0) and (2,0).

Rather, as said earlier, if you verify that S={p1, p2, ..., p5} is linearly independent (just as you did in part (a)), then this set forms a basis for the set of polynomials, so span(S) = the whole set and so certainly p belongs to span(S).

 

[billli]

Thanks for the reply Unco!
from matlab:
http://img156.imageshack.us/img156/582/set.jpg
so P(x) belongs to the span of S. as it has a unique solution.

 

[Unco]

...so P(x) belongs to the span of S. as it has a unique solution.

The first 5 columns of your row-reduced matrix show that S forms a basis. This is enough.

The last (sixth) column does give you the coefficients of the linear combination of p1, ..., p5 giving p.

The entire 5x6 matrix is the augmented matrix for the system

$$[\mathbf{p_1} \, \ldots \, \mathbf{p_5}][k_1 \, \ldots \, k_5]^T = \mathbf{p},$$

where $$k_i$$ is the coefficient of $$\mathbf{p_i}$$ in such a linear combination, and one wishes to solve for $$[k_1 \, \ldots \, k_5]^T$$.

The point is that not all the coefficients of the linear combination, i.e., the k_i, are non-zero.